# Power and Product Rule

• June 2nd 2009, 11:21 AM
lisa1984wilson
Power and Product Rule
Differentiate the given function (two methods)

I only know how to do the chain rule which I get 8(4x-7)

please show how to do this so I can master the rest of the questions my self..thanks
• June 2nd 2009, 11:22 AM
Amer
• June 2nd 2009, 11:30 AM
lisa1984wilson
Sorry I can't believe I did not post it:

(7-4x)^2
• June 2nd 2009, 11:37 AM
Amer
you differentiate it correct there is a negative sign

$f(x)=(7-4x)^2$

$f'(x)=2(-4)(7-4x)$

$f(x)=(-3x^2)^{n}....f'(x)=n(-3x^2)'(-3x^2)^{n-1}=n(-3(2)(x)^{2-1})(-3x^2)^{n-1}$
• June 2nd 2009, 11:43 AM
lisa1984wilson
How do I show that the two answers are the same (product rule and power rule) Thanks
• June 2nd 2009, 11:49 AM
Amer
$f(x)=(7-4x)(7-4x)$

I will differentiate it by product rule

$f'(x)=(7-4x)(-4)+(-4)(7-4x)=2(-4)(7-4x)$

in general

$f(x)=AB...A,B..functions..of..x$

differentiate

$f'(x)=A\frac{d}{dx}(B)+B\frac{d}{dx}(A)$
• June 2nd 2009, 02:56 PM
lisa1984wilson
What you did for the power rule..I don't understand it! Is that all the steps for it? I got the product rule that is pretty easy thanks!
• June 2nd 2009, 03:00 PM
craig
Quote:

Originally Posted by lisa1984wilson
What you did for the power rule..I don't understand it! Is that all the steps for it? I got the product rule that is pretty easy thanks!

Quote:

Originally Posted by Amer
you differentiate it correct there is a negative sign

$f(x)=(7-4x)^2$

$f'(x)=2(-4)(7-4x)$

$f(x)=(-3x^2)^{n}....f'(x)=n(-3x^2)'(-3x^2)^{n-1}=n(-3(2)(x)^{2-1})(-3x^2)^{n-1}$

That is the product rule, the second equation is the differentiated form of your equation, and in the third Amer is showing you how to do it for any equation.

You take your original equation, reduce the power by 1, multiply the whole thing by the original power, this is the $2$ in your equation, and then multiply the whole thing by the derivative of what's in the brackets, in your equation this is $(-4)$.

Hope this helps
• June 2nd 2009, 03:04 PM
lisa1984wilson
I know the product rule its the power rule which I don't understand
• June 2nd 2009, 03:08 PM
craig
Quote:

Originally Posted by lisa1984wilson
I know the product rule its the power rule which I don't understand

If you have an equation $y = (f(x))^n$.

Then $\frac{dy}{dx} = nf'(x)(f(x))^{n-1}$

As I said:

Quote:

Originally Posted by craig
You take your original equation, reduce the power by 1, multiply the whole thing by the original power, this is the $2$ in your equation, and then multiply the whole thing by the derivative of what's in the brackets, in your equation this is $(-4)$.

• June 2nd 2009, 03:10 PM
lisa1984wilson
sorry I misunderstood...I've been working on calculus all day getting tired...thanks!
• June 2nd 2009, 03:11 PM
Plato
Quote:

Originally Posted by lisa1984wilson
I know the product rule its the power rule which I don't understand

$\begin{gathered}
\frac{{df^n }}
{{dx}} = n \cdot f^{n - 1} \cdot f' \hfill \\
\frac{{d\left( {x^3 + x} \right)^7 }}
{{dx}} = 7\left( {x^3 + x} \right)^6 \left( {3x^2 + 1} \right) \hfill \\
\end{gathered}$
• June 2nd 2009, 03:11 PM
craig
Quote:

Originally Posted by lisa1984wilson
sorry I misunderstood...I've been working on calculus all day getting tired...thanks!

No worries, hope it's all cleared up now :)